Exponential and Logarithmic Equations


Contents: This page corresponds to § 4.4 (p. 348) of the text.

Suggested problems from text:

p. 355 #13,17,21,25,27,29,31,51,53,57,59,81,89

Exponential Equations

Logarithmic Equations


Exponential Equations

Some exponential equations can be solved by using the fact that exponential functions are one-to-one. In other words, an exponential function does not take two different values to the same number.

Example 1.

3x = 9

3x = 32

The function f(x) = 3x is one-to-one, so it does not take two different values to 9, so x must equal 2.

x = 2

The equation in example 1 was easy to solve because we could express 9 as a power of 3. However, it is often necessary to use a logarithm when solving an exponential equation.


Example 2.

ex = 20

We are going to use the fact that the natural logarithm is the inverse of the exponential function, so ln ex = x, by logarithmic identity 1. We must take the natural logarithm of both sides of the equation.

ln ex = ln 20

Now the left hand side simplifies to x, and the right hand side is a number. It is approximately 2.9957.

x = 2.9957

Exercise 1:

Use a calculator to check the answer we found to the equation in example 2.

Example 3.

5x = 16 We will solve this equation in two different ways.

First Approach: We use the fact that log5 5x = x (logarithmic identity 1 again).

5x = 16

log5 5x = log5 16

x = log5 16

x = ln 16 / ln 5, by the change-of-base formula.

x = 1.7227 (approximately)

Second Approach: We will use the natural logarithm and property 3.

5x = 16 Take the natural logarithm of both sides.

ln 5x = ln 16

x ln 5 = ln 16

x = ln 16 / ln 5

x = 1.7227 (approximately)

We could have used any logarithm with the second approach. The second approach is the one that you see most often.

Exercise 2:

Use a calculator to check the answer we found to the equation in example 3.

Equations like that in the next example occur frequently in applications.

Example 4.

200 e0.07t = 500

We first isolate the exponential part by dividing both sides of the equation by 200.

e0.07t = 2.5

Now we take the natural logarithm of both sides.

ln e0.07t = ln 2.5

The left hand side simplifies to 0.07t, by logarithmic identity 1.

0.07t = ln 2.5

t = ln (2.5) / 0.07

t = 13.1 (approximately)

Exercise 3:

Solve the following equations and check the answers.

(a) 3x = 10

(b) 150 e0.05 t = 350

Answer

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Logarithmic Equations

When solving exponential equations we frequently used logarithmic identity 1 because it involves applying a logarithmic function to "undo" the effect of an exponential function. When dealing with logarithmic equations we will use logarithmic identity 2 where an exponential function is applied to "undo" the effect of a logarithmic function.

Example 5.

2 log x = 12

We want to isolate the log x, so we divide both sides by 2.

log x = 6

Since log is the logarithm base 10, we apply the exponential function base 10 to both sides of the equation.

10log x = 106

By logarithmic identity 2, the left hand side simplifies to x.

x = 106 = 1000000

Example 6.

7 + 3 ln x = 15 First isolate ln x.

3 ln x = 8

ln x = 8/3

Now apply the exponential function to both sides.

eln x = e8/3

x = e8/3

This is the exact answer. If you use a calculator to evaluate this expression, you will have an approximation to the answer.

x is approximately equal to 14.39.

Exercise 4:

Check the answers found in examples 5 and 6.

Example 7.

ln (x + 4) + ln (x - 2) = ln 7

First we use property 1 of logarithms to combine the terms on the left.

ln (x + 4)(x - 2) = ln 7

Now apply the exponential function to both sides.

eln (x + 4)(x - 2) = eln 7

The logarithmic identity 2 allows us to simplify both sides.

(x + 4)(x - 2) = 7

x2 + 2x - 8 = 7

x2 + 2x - 15 = 0

(x - 3)(x + 5) = 0

x = 3 or x = -5

x = 3 checks, for ln 7 + ln 1 = ln 7.

x = -5 does not check, for when we try to substitute -5 for x in the original equation we are taking the natural logarithm of negative numbers, which is not defined.

So, x = 3 is the only solution.

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